Problem: The lifespans of seals in a particular zoo are normally distributed. The average seal lives $15$ years; the standard deviation is $3.3$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a seal living between $8.4$ and $24.9$ years.
Solution: $15$ $11.7$ $18.3$ $8.4$ $21.6$ $5.1$ $24.9$ $99.7\%$ $95\%$ $2.35\%$ $2.35\%$ We know the lifespans are normally distributed with an average lifespan of $15$ years. We know the standard deviation is $3.3$ years, so one standard deviation below the mean is $11.7$ years and one standard deviation above the mean is $18.3$ years. Two standard deviations below the mean is $8.4$ years and two standard deviations above the mean is $21.6$ years. Three standard deviations below the mean is $5.1$ years and three standard deviations above the mean is $24.9$ years. We are interested in the probability of a seal living between $8.4$ and $24.9$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the seals will have lifespans within 3 standard deviations of the average lifespan. It also tells us that $95\%$ of the seals will have lifespans within 2 standard deviations of the mean. That leaves $99.7\% - 95\% = 4.7\%$ of seals between 2 and 3 standard deviations of the mean, or $2.35\%$ on either side of the distribution. The probability of a particular seal living between $8.4$ and $24.9$ years is ${95\%} + \color{orange}{2.35\%}$, or $97.35\%$.